I have a function to calculate some weighted arithmetic mean of a vector $x\in \mathbb{R}^N$ $$f(x) = \frac{\sum\limits_i^N w(i)^2x_i}{\sum\limits_i^N w(i)}$$
Now for $w(i) = 1$, this simply resolves to $\frac{1}{N}\sum\limits_i^N x_i = \bar{x}$, i.e. the regular, arithmetic mean.
But suppose we define $w(i) = \frac{1}{i^2}$. Is there a way to correct f(x) by some factor to get back to the (not weighted) arithmetic mean? So is there an $a$, such that $$af(x) = \bar{x}$$
This might be quite easy, or impossible for some reason. But I can't figure it out.
No, it is not possible. To show why, it suffices to choose two distinct samples which have the same weighted mean, but different unweighted means, immediately implying that there is no way to undo the weighting. That is to say, we consider $$\boldsymbol x = (x_1, \ldots, x_N), \\ \boldsymbol y = (y_1, \ldots, y_N)$$ such that $$f(\boldsymbol x) = f(\boldsymbol y)$$ but $$\bar x \ne \bar y.$$ These criteria comprise a linear system of two equations in $N$ unknowns (for fixed weighting factors $w(i)$). Note that the second condition is not strictly an equation, but can be converted into one by postulating, for instance, $\bar x = \bar y + k$ for some nonzero constant $k$. With $N$ larger than $2$, or with $N$ sufficiently large, this system is in general underdetermined, thus will admit infinitely many solutions.